Q.

What volume of 90% alcohol by weight (d  =0.8 g ml-1) must be used to prepare 80 mL of 16% alcohol by weight (d =0.9 g ml-1)?

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answer is 10.

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Detailed Solution

Use M=% by weight x 10 xdMw2M1V1=M2V290×10×0.846×V=10×16×0.946×80V=16×0.9×8090×0.8=16mL
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What volume of 90% alcohol by weight (d  =0.8 g ml-1) must be used to prepare 80 mL of 16% alcohol by weight (d =0.9 g ml-1)?